Why do Fresnel-integrals contain $sqrt{pi}$?

The non-elementary functions

$$ F(x) = int sin(x^2)mathrm dx $$
$$ G(x) = int cos(x^2)mathrm dx $$

will yield

$$ F(x) =sum_{k=1}^{infty} (-1)^k frac{x^{(4k+3)}}{(2k+1)!(4k+3)}$$
$$ G(x) =sum_{k=1}^{infty} (-1)^k frac{x^{(4k+1)}}{(2k)!(4k+1)} $$

respectively, If you simply rewrite it to a Taylor series, and then integrate every term

However, some places (for example integral-calulator.com) claim that the integral is equal to

$$ F(x) = sqrt{frac{pi}{2}} SBig(sqrt{frac{2}{pi}} xBig) + C $$
$$ G(x) = sqrt{frac{pi}{2}} CBig(sqrt{frac{2}{pi}} xBig) + C $$
where
$$S(x) =sum_{k=1}^{infty} (-1)^k frac{x^{(4k+3)}}{(2k+1)!(4k+3)} $$
$$C(x) =sum_{k=1}^{infty} (-1)^k frac{x^{(4k+1)}}{(2k)!(4k+1)} $$

Where does the $ sqrt{frac{pi}{2}} $ and $ sqrt{frac{2}{pi}}$ term come from? Why can’t one just say that $ F(x) = S(x) + C $ and $ G(x) = C(x) + C $?